Vector A points in the negative y direction and has a magnitude of 6 units. Vector B has twice the magnitude and points in the positive x direction.

Find the direction and magnitude of vec A-B and Vec B-A?

a 150 lb skydiver is falling in a head first position at a terminal (constant) velocity of 120 mi/h find the force in pounds of the skydiver

To find the direction and magnitude of vector A-B and vector B-A, we can break them down into their components.

Given:
Vector A points in the negative y direction and has a magnitude of 6 units.
Vector B has twice the magnitude of vector A and points in the positive x direction.

Let's start by finding the components of vector A and vector B.

Vector A:
Magnitude: 6 units
Direction: Negative y direction

The components of vector A can be calculated as:
Ax = 0 (since vector A does not have an x-component)
Ay = -6

Vector B:
Magnitude: 2 * magnitude of vector A = 2 * 6 = 12 units
Direction: Positive x direction

The components of vector B can be calculated as:
Bx = 12
By = 0 (since vector B does not have a y-component)

Now, let's calculate the components of vector A-B and vector B-A.

Components of vector A-B:
(A-B)x = Ax - Bx = 0 - 12 = -12
(A-B)y = Ay - By = -6 - 0 = -6

Components of vector B-A:
(B-A)x = Bx - Ax = 12 - 0 = 12
(B-A)y = By - Ay = 0 - (-6) = 6

Now that we have the components, we can find the magnitude and direction of vector A-B and vector B-A.

Magnitude of vector A-B:
|(A-B)| = sqrt((A-B)x^2 + (A-B)y^2) = sqrt((-12)^2 + (-6)^2) = sqrt(144 + 36) = sqrt(180) = 12√5 units

Direction of vector A-B:
The direction of vector A-B can be found using the arctan function:
θ = arctan((A-B)y / (A-B)x) = arctan((-6) / (-12)) = arctan(0.5)
θ ≈ 26.57 degrees

Magnitude of vector B-A:
|(B-A)| = sqrt((B-A)x^2 + (B-A)y^2) = sqrt((12)^2 + (6)^2) = sqrt(144 + 36) = sqrt(180) = 12√5 units

Direction of vector B-A:
The direction of vector B-A can be found using the arctan function:
θ = arctan((B-A)y / (B-A)x) = arctan(6 / 12) = arctan(0.5)
θ ≈ 26.57 degrees

Therefore, the magnitude of vector A-B and vector B-A is 12√5 units, and the direction of both vectors is approximately 26.57 degrees.

To find the direction and magnitude of vector A-B, we need to find the resulting vector obtained by subtracting vector B from vector A. Similarly, to find the direction and magnitude of vector B-A, we need to subtract vector A from vector B.

Let's first find the resulting vector A-B:

1. Draw a diagram to visualize the vectors A and B.
- Vector A points in the negative y direction with a magnitude of 6 units. So, draw a vector pointing downward with a length of 6 units.
- Vector B has twice the magnitude of A and points in the positive x direction. Draw a vector pointing rightward with a length of 2 * 6 = 12 units.

2. To compute vector A-B, subtract the components of vector B from vector A.
- Subtract the x-components: 0 - 12 = -12.
- Subtract the y-components: -6 - 0 = -6.

3. The resulting vector A-B has a magnitude and direction relative to the origin.
- The magnitude (length) of vector A-B can be found using the Pythagorean theorem: magnitude = sqrt((-12)^2 + (-6)^2) ≈ 13.42 units.
- The direction of vector A-B can be found using trigonometry. The angle can be found as arctan((-6)/(-12)), which is approximately equal to 26.57 degrees. However, since vector A-B has negative components, the angle is in the fourth quadrant: 180 degrees + 26.57 degrees ≈ 206.57 degrees.

So, the magnitude of vector A-B is approximately 13.42 units, and its direction is approximately 206.57 degrees counterclockwise from the positive x-axis.

Next, let's find the resulting vector B-A:

1. Again, draw a diagram to visualize the vectors A and B.

2. To compute vector B-A, subtract the components of vector A from vector B.
- Subtract the x-components: 12 - 0 = 12.
- Subtract the y-components: 0 - (-6) = 6.

3. The resulting vector B-A has a magnitude and direction relative to the origin.
- The magnitude (length) of vector B-A can be found using the Pythagorean theorem: magnitude = sqrt(12^2 + 6^2) = 12.81 units.
- The direction of vector B-A can be found using trigonometry. The angle can be found as arctan(6/12), which is approximately equal to 26.57 degrees. Since vector B-A has positive components, the angle is in the first quadrant: 26.57 degrees.

So, the magnitude of vector B-A is approximately 12.81 units, and its direction is approximately 26.57 degrees counterclockwise from the positive x-axis.